The world's growth and decay stories all wear the same two outfits. A bacterial colony doubling in a petri dish, savings quietly compounding in an account, a radioactive sample ticking down atom by atom, a cup of coffee losing its heat to the room — wildly different stories, identical mathematical wardrobe. Whenever the change in a quantity is proportional to the quantity itself — the more you have, the faster it grows (or shrinks) — the quantity follows an exponential model. And whenever you turn the question around and ask "how long until…?" — how long until the money doubles, the caffeine wears off, the sample halves — you need the exponential's inverse: a logarithm.

Modelling, then, is a two-step craft: choose the outfit (spot that the situation is exponential), and fit the buttons (pin down the two numbers that tailor the general formula to your story). This page teaches both — and how logs pull the unknown down out of the exponent when the question is about time.

The model, and how to read it

Many real-world quantities change by the same factor in every equal step of time: a population multiplies, savings earn interest, a radioactive sample halves. Each is an exponential model:

Here A is the starting amount (the value when t = 0), k is the growth or decay factor per unit of time, and t is the time. This is the same multiply-by-a-constant pattern you have seen in , now read off a smooth curve instead of a list of terms.

Two numbers tell the whole story. Set t = 0: since k^0 = 1, the value is A — the starting amount. Then each extra unit of time multiplies by k:

The whole drama of "growth versus decay" lives in a single comparison — k against 1:

• Growth when k > 1 — the value rises, and rises ever faster (a doubling population has k = 2).
• Decay when 0 < k < 1 — the value falls, but never quite reaches zero (a sample that halves has k = \tfrac{1}{2}).
• At exactly k = 1 nothing happens at all: y = A forever. The boundary between the two worlds is perfectly flat.

A percentage rate folds neatly into k. Growing by 5% a year means "keep your 100% and add 5%", so k = 1 + 0.05 = 1.05. Losing 5% a year means "keep 95%", so k = 1 - 0.05 = 0.95. This translation — rate to factor — is the single most common button-fitting move in the whole subject, and it is where these become a formula you can compute with.

See it: growth versus decay

Slide A to set where the curve starts, and slide k across 1. Above 1 the curve climbs — gently at first, then steeply, because each step multiplies an ever-larger amount. Below 1 it sinks toward zero without ever touching it — each step removes the same fraction, but a fraction of a shrinking amount is a shrinking removal. At exactly 1 it is flat. Time only runs forward here, so t \ge 0.

Try this: set k = 2 and pick any starting height on the curve. However high or low you start, the curve takes exactly one unit of time to double from there. That is the exponential's signature — a constant doubling time for growth, a constant half-life for decay — and it is precisely the property that a straight line does not have. If a data set doubles in equal time steps, it is wearing the exponential outfit.

Using logs to find the time

Evaluating a model is easy: plug in t, get y. The interesting questions run the other way — you know the target value and want the time. Now t sits up in the exponent, and the tool that reaches up and pulls it down is the logarithm. To find when a colony of A = 200 bacteria that doubles each hour (k = 2) reaches 1600:

The recipe is always the same three moves: divide by the starting amount to isolate k^t, take a log of both sides to bring the exponent down, then divide by \log k. In general:

Any log base works — \log_{10}, \ln, your calculator's choice — because the base cancels in the ratio. The same recipe finds a half-life, a doubling time, or how long savings take to reach a goal. Let's use it on real money.

Worked example: when does money double?

You invest P pounds at 5% compound interest per year. First, fit the buttons: 5% growth per year means the factor is k = 1.05, and the starting amount is P, so the model is

Question: how long until the money doubles? We want the time t when A(t) = 2P. Follow the recipe, step by step:

Step 1 — divide by the starting amount. The P cancels completely — the doubling time doesn't depend on how much you invested:

Step 2 — take a log of both sides and use the power rule \log(1.05^{\,t}) = t\log 1.05:

Step 3 — divide to leave t alone:

So at 5% a year, money doubles in a little over 14 years — whether you started with £10 or £10 million. Notice the shape of the answer: t = \log 2 / \log k is the doubling-time formula, and swapping the 2 for \tfrac{1}{2} (or 3, or 10) gives the half-life, tripling time, or ten-folding time the same way. Bankers, incidentally, get to roughly the same 14.2 without a calculator — see the vignette below.

Worked example: the caffeine clock

Decay models often come pre-buttoned in half-life form. Caffeine leaves the body with a half-life of roughly 6 hours: whatever amount is in your blood now, half of it will be gone 6 hours from now. A cup of coffee delivering C_0 = 80 mg is modelled by

Read the exponent carefully: t/6 counts how many half-lives have passed. At t = 6 the exponent is 1 (one halving); at t = 12 it is 2 (two halvings, a quarter left). Writing the model this way saves you converting the half-life into a per-hour factor — though you could: k = (1/2)^{1/6} \approx 0.891, an 11% loss per hour.

Question: how much caffeine is left 15 hours after an 80 mg coffee? Fifteen hours is 15/6 = 2.5 half-lives:

About 14 mg — an evening espresso is still very much on duty at 4 a.m. And the inverse question uses the same log recipe as the money example: "when is only 10 mg left?" means (1/2)^{t/6} = 10/80 = 1/8 = (1/2)^3, so t/6 = 3 and t = 18 hours. When the numbers are less friendly, t = 6\log(C/C_0)/\log(1/2) does the job.

Worked example: fitting the model to data

So far the buttons were handed to us. In real modelling you get data and must find A and k yourself. Two data points are exactly enough, because there are exactly two unknowns.

A biologist counts a bacterial culture twice: N(0) = 500 cells at the start, and N(3) = 4000 cells three hours later. Fit N(t) = A\,k^{\,t}.

Step 1 — the starting value gives A for free:

Step 2 — substitute the second point and solve for k:

Notice the division in the middle: k^3 = N(3)/N(0). The starting amount always cancels out of the ratio, so the factor comes purely from how much the quantity multiplied between the two measurements. The fitted model is N(t) = 500 \cdot 2^{\,t} — the culture doubles every hour.

If the ratio hadn't been a friendly perfect cube — say k^3 = 5 — the root is still routine: k = 5^{1/3} \approx 1.710, or with logs, \log k = \tfrac{1}{3}\log 5. The general recipe for two points (0, N_0) and (T, N_T):

And that is the whole craft in one line: one measurement pins the start, the ratio of two measurements pins the factor, and logs answer every "when?" that follows.

Khan Academy works through exponential model word problems here: